The question provides the following parameters:
![\begin{gathered} \mu=16.3 \\ \sigma=4.2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4pshec1q6s5kqkkecdsfo608iuxw3dgtos.png)
For 15 minutes, the z-score is calculated using the formula:
![z=(x-\mu)/(\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/h06hsre30elxbqnbdkqzw5pbp57988qa0r.png)
At x = 15:
![z=(15-16.3)/(4.2)=-0.3](https://img.qammunity.org/2023/formulas/mathematics/college/anndd1qatxq158n04pcwazswbd1st9u06l.png)
The probability is calculated using the formula:
![P(X<15)=Pr(z<-0.3)=Pr(z<0)-Pr(0From tables, we have:[tex]\begin{gathered} Pr(z<0)=0.5 \\ Pr(0Therefore, the probability is given to be:[tex]\begin{gathered} P(X<15)=0.5-0.1179 \\ P(X<15)=0.38 \end{gathered}]()
The expected number of callers will be calculated using the formula:
![\begin{gathered} E=xP(x) \\ At\text{ }x=1500 \\ E=1500*0.38 \\ E=570 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hutvg45pp2kp8hz5dm9yvmtl0xybqkyxr9.png)
Therefore, the expected number of callers whose calls last less than 15 minutes is 570 callers.